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Schaum's Chapter 16 Heat Quantities

Schaum's Chapter 16 Heat Quantities 1. How much heat is required to raise the temperature of 250 ml of water from 20 0 C to 35 0 C? 2. How much heat does 25 g of aluminum give off as it cools from 100 0 C to 20 0 C?

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Schaum's Chapter 16 Heat Quantities

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  1. Schaum's Chapter 16 Heat Quantities 1. How much heat is required to raise the temperature of 250 ml of water from 200 C to 350C? 2. How much heat does 25 g of aluminum give off as it cools from 1000C to 200 C? 3. A certain amount of heat is added to a mass of aluminum and its temperature is raised 570C. Suppose the same amount of heat is added to the same mass of copper. How much does the temperature of copper rise? 21. Furnace oil has a heat of combustion of 44 MJ/kg. Assuming that 70% of the heat is useful, how many kilograms of oil are required to raise the temperature of 2000 kg of water from 200C to 990C? 250 ml = 250g = .25kg c = 4186 j/kg C ΔT = (35 - 20) = 15 Q = .25(4186)(15) Q = 15697.5 J = 15.7 kJ Q = mcΔT 25g =.025kg c = 880 j/kg C ΔT = (20 - 100) = -80 Q = .025(880)(-80) Q = - 1760 J = - 1.76 kJ Q = mcΔT +Q heat gain (heat enters an object) -Q heat loss (heat leaves an object) Q = mcΔT Q AL = mcΔT Q AL = m(880)(57) Q AL = 50160m Q Cu = mcΔT 50160m = m(390)(ΔT) ΔT = 128.62 0C useful Q = 44 MJ (.70) =30.8 MJ Q = mcΔT Q = 2000(4186)(79) Q = 661388000 J = 661.39 MJ 661.39 MJ( 1 kg/ 30.8 MJ) = 21.47 kg

  2. 5. A thermos bottle contains 250 g of coffee at 900C. To this is added 20g of milk at 50C. After equilibrium is established, what is the temperature of the liquid? 6. A thermos bottle contains 150 g of water at 40C. Into this is placed 90g of metal at 1000C. After equilibrium is established, the temperature of the water and metal is 210C. What is the specific heat capacity of the metal? THIS IS THE LAB!!!! 7. A 200g copper calorimeter can contains 150 g of oil at 200C. To the oil is added 80 g of aluminum at 3000C. What will be the temperature of the system after equilibrium is established? 8. Exactly 3 g of carbon was burned to CO2 in a copper calorimeter. The mass of the calorimeter is 1500 g and there is 2000g of water in the calorimeter. The initial temperature was 200 C and the final temperature is 310 C. Calculate the heat given off per gram of carbon as it burns, (mcΔT)coffee = (mcΔT)milk Qlost = Qgained All have c = 4186 J/kgC .25(4186)(90 - T2)= .02(4186)( T2 - 5) T2 = 83.67 0C (mcΔT)coffee = (mcΔT)milk (mcΔT)metal = (mcΔT)water Qlost = Qgained .09c(100 - 21) = .15(4186)(21-4) c = 1501.31 J/kg C (mcΔT)metal = (mcΔT)water Qlost = Qgained (mcΔT)AL= (mcΔT)Cu + (mcΔT)oil .08(900)(300 - T2) = [.2(390) + .150(1549)](T2 - 20) T2 = 72.73 0C Qc = 23534.5 cal = 23.53 kcal Qlost = Qgained (mcΔT)C= (mcΔT)Cu + (mcΔT)water (mcΔT)C= [1.5(93) + 2(1000)](31 - 20) Qc = 23534.5 cal 23.53 kcal / 3 g = 7.84 kcal / g

  3. 9. Determine the temperature T2 that results when 150 g of ice at 00C is mixed with 300 g of water at 500C. 10. How much heat is given up when 20 g of steam at 1000C is condensed and cooled to 200C? 11. A 20 g piece of aluminum at 900C is dropped into a cavity in a large block of ice at 00C. How much ice melts? Qlost = Qgained (mcΔT)water = ice to turn to water + water to change temperature .3(4186)(50 - T2) = .15(3.33 x 105) + .15 (4186) (T2-0) T2 = 6.82 0C Steam change to water + water cool to 20 .02(22.6 x 105) + .02(4186)(20 - 100) Q = - 51897.6 = 51.9 kJ +Q heat gain (heat enters an object) -Q heat loss (heat leaves an object) Qlost = Qgained (mcΔT)Al = ice to turn to water (melts) .02(880)(90 - 0) = m(3.33 x 105) m = .004757 kg = 4.76 g

  4. 13. An electric heater that produces 900 W of power is used to vaporize water. How much water at 100 0C can be changed to steam at 1000C in 3 minutes by the heater? 14. A 3 gram bullet (c = 128 J/kgC) moving at 180 m/s enters a bag of sand and stops. By what amount does the temperature of the bullet change if all of its KE becomes thermal energy that is added to the bullet? 15. Suppose a 60kg person consumes 2500 Cal of food in one day. If the entire heat equivalent of this food were retained by the person's body, how large a temperature change would it cause? For the body c = .83 cal/gC 1 Cal = 1 kcal = 1000 cal Q = mLv 3 min (60 sec/min) = 180 s P = W/t = Q/t 900 W = 900 J/s 162000 = m (22.6 x 105) m = 0.07168 kg = 71.68 g 900 J/s ( 180 s) = 162000 J 1/2 mv2 = mcΔT 1/2 (.003)(180)2 = (.003)(128)(ΔT) ΔT = 126.56 0C ΔKE = ΔQ 1/2 mv2 = mcΔT Q = mcΔT 2500000 = 60000(.83)ΔT ΔT = 50.200C Q = 2500 Cal ( 1000 cal/Cal) = 2500000 cal m = 60 kg = 60000g

  5. 27. How much heat is required to change 10 g of ice at 00C to steam at 1000C? ice at MP 00 C Change state Q = mLf Q = .01(3.33 x 105) Q = 3330 J water at 00 C raise temperature to BP 100 Q = mcΔT Q = .01(4186)(100 - 0) Q = 4186 J Q = 3330 + 4186 + 22600 = 30116 J = 30.12 kJ 100 steam at BP 1000 C Change state Q = mLv Q = .01(22.6 x 105) Q = 22600 J 22.6 kJ 7.52 + 22.6 30.12 kJ 0 C water at 00 C raise temperature to BP 100 Q = mcΔT Q = .01(4186)(100 - 0) Q = 4186 J 4.19 kJ 3.33 + 4.19 7.52 kJ 0 ice at MP 00 C Change state Q = mLf Q = .01(3.33 x 105) Q = 3330 J 3.33 kJ 30.12 7.52 3.33 kJ

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